3626: [LNOI2014]LCA (树链剖分+离线处理)

最新推荐文章于 2022-03-16 23:20:55 发布

ndsffx501ccy

最新推荐文章于 2022-03-16 23:20:55 发布

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分类专栏：数据结构其他

本文链接：https://blog.youkuaiyun.com/ndsffx501ccy/article/details/38264689

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本文介绍了一个结合线段树和树状数组的数据结构实现，通过具体实例展示了如何利用这些高级数据结构解决复杂查询与更新问题。文中详细解释了算法流程，包括节点插入、深度优先搜索、树链剖分等关键步骤，并实现了高效的区间更新和查询操作。

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#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x*f;
}
const int maxn = 50005, inf = 1000000000, mod = 201314;
int n, m, tot, cnt, place, son[maxn], deep[maxn], head[maxn], fa[maxn], belong[maxn], pl[maxn];

struct edge {
    int to, next;
} e[maxn];

struct que {
    int z, ans1, ans2;
} q[maxn];

struct data {
    int num, p;
    bool flag;
} a[maxn << 1];

struct seg {
    int l, r, sum, tag;
} tr[maxn << 2];

inline bool operator<(data a, data b) {
    return a.p < b.p;
}

inline void insert(int u, int v) {
    e[++cnt] = (edge){v, head[u]};
    head[u] = cnt;
}

inline void dfs1(int x) {
    son[x] = 1;
    for (int i = head[x]; i; i = e[i].next) {
        if (e[i].to == fa[x])continue;
        deep[e[i].to] = deep[x] + 1;
        fa[e[i].to] = x;
        dfs1(e[i].to);
        son[x] += son[e[i].to];
    }
}

inline void dfs2(int x, int chain) {
    pl[x] = ++place;
    belong[x] = chain;
    int k = n;
    for (int i = head[x]; i; i = e[i].next)
        if (e[i].to != fa[x] && son[e[i].to] > son[k])
            k = e[i].to;
    if (k != n)dfs2(e[k].to, chain);
    for (int i = head[x]; i; i = e[i].next)
        if (e[i].to != fa[x] && e[i].to != k)
            dfs2(e[i].to, e[i].to);
}

inline void pushdown(int k) {
    if (tr[k].l == tr[k].r || !tr[k].tag)return;
    int tag = tr[k].tag;
    tr[k].tag = 0;
    tr[k << 1].sum += (tr[k << 1].r - tr[k << 1].l + 1) * tag;
    tr[k << 1 | 1].sum += (tr[k << 1 | 1].r - tr[k << 1 | 1].l + 1) * tag;
    tr[k << 1].tag += tag;
    tr[k << 1 | 1].tag += tag;
}

inline void build(int k, int l, int r) {
    tr[k].l = l;
    tr[k].r = r;
    if (l == r)return;
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
}

inline void update(int k, int x, int y) {
    pushdown(k);
    int l = tr[k].l, r = tr[k].r;
    if (x == l && y == r) {
        tr[k].tag++;
        tr[k].sum += r - l + 1;
        return;
    }
    int mid = (l + r) >> 1;
    if (y <= mid)update(k << 1, x, y);
    else if (x > mid)update(k << 1 | 1, x, y);
    else update(k << 1, x, mid), update(k << 1 | 1, mid + 1, y);
    tr[k].sum = tr[k << 1].sum + tr[k << 1 | 1].sum;
}

inline void solve_update(int x, int f) {
    while (belong[x] != belong[f]) {
        update(1, pl[belong[x]], pl[x]);
        x = fa[belong[x]];
    }
    update(1, pl[f], pl[x]);
}

inline int query(int k, int x, int y) {
    pushdown(k);
    int l = tr[k].l, r = tr[k].r;
    if (l == x && r == y)return tr[k].sum;
    int mid = (l + r) >> 1;
    if (y <= mid)return query(k << 1, x, y);
    else if (x > mid)return query(k << 1 | 1, x, y);
    else return query(k << 1, x, mid) + query(k << 1 | 1, mid + 1, y);
}

inline int solve_query(int x, int f) {
    int res = 0;
    while (belong[x] != belong[f]) {
        res += query(1, pl[belong[x]], pl[x]);
        res %= mod;
        x = fa[belong[x]];
    }
    res += query(1, pl[f], pl[x]);
    res %= mod;
    return res;
}

int main() {
    n = read();
    m = read();
    for (int i = 1; i < n; i++) {
        int x = read();
        insert(x, i);
    }
    for (int i = 1; i <= m; i++) {
        int l = read(), r = read();
        q[i].z = read();
        a[++tot].p = l - 1;
        a[tot].num = i;
        a[tot].flag = 0;
        a[++tot].p = r;
        a[tot].num = i;
        a[tot].flag = 1;
    }
    build(1, 1, n);
    sort(a + 1, a + tot + 1);
    dfs1(0);
    dfs2(0, 0);
    int now = -1;
    for (int i = 1; i <= tot; i++) {
        while (now < a[i].p)solve_update(++now, 0);
        int t = a[i].num;
        if (!a[i].flag)q[t].ans1 = solve_query(q[t].z, 0);
        else q[t].ans2 = solve_query(q[t].z, 0);
    }
    for (int i = 1; i <= m; i++)
        printf("%d\n", (q[i].ans2 - q[i].ans1 + mod) % mod);
    return 0;
}